[USACO08FEB]修路Making the Grade

Posted 2021-12-30 excellent-zzy

　　给个n <= 2000长度数列，可以把每个数改为另一个数代价是两数之差的绝对值。求把它改为单调不增or不减序列最小代价。

　　把高度离散之后DP。。。存在b数组中b[j]表示第j大的高度。

我们用f[i][j]将前i段变作不下降序列，且第j段道路的高度为b[j]时的最小花费，显而易见，

　　f[i][j] = min(f[i - 1][k]) + abs(a[i] - b[j])(1<=k<=j）其中a[i]表示第i段路原本的高度。

　　枚举k的话，你会发现时间复杂度为n^3。为了解决这个问题，我们发现min(f[i - 1][k])是可以在做第i - 1段路段的时候处理出来的，所以复杂度就成了n ^ 2。

#define B cout << "BreakPoint" << endl;
#define O(x) cout << #x << " " << x << endl;
#define O_(x) cout << #x << " " << x << " ";
#define Msz(x) cout << "Sizeof " << #x << " " << sizeof(x)/1024/1024 << " MB" << endl;
#include<cstdio>
#include<cmath>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#define LL long long
#define inf 1000000009
#define N 2019
using namespace std;
int n,m,c[N][N],f[N][N],a[N],b[N],t[N],ans;
inline int read() 
	int s = 0,w = 1;
	char ch = getchar();
	while(ch < ‘0‘ || ch > ‘9‘) 
		if(ch == ‘-‘)
			w = -1;
		ch = getchar();
	
	while(ch >= ‘0‘ && ch <= ‘9‘) 
		s = s * 10 + ch - ‘0‘;
		ch = getchar();
	
	return s * w;

void ini() 
	for (int i = 0; i <= n; i++) 
		for (int j = 0; j <= m; j++) 
			c[i][j] = f[i][j] = 0;
		
	
	return ;

void dp() 
	for (int i = 1; i <= n; i++) 
		for (int j = 1; j <= m; j++) 
			f[i][j] = c[i - 1][j] + abs(a[i] - b[j]);
			if (j != 1) c[i][j] = min(c[i][j - 1],f[i][j]);
			else c[i][j] = f[i][j];
		
	
	return ;

void init() 
	n = read();
	for(int i = 1; i <= n; i++) 
		a[i] = read();
		t[i] = a[i];
	
	return ;

void solve() 
	sort(t + 1,t + 1 + n);
	int res = -1;
	for(int i = 1; i <= n; i++) 
		if(res != t[i]) 
			b[++m] = t[i];
			res = t[i];
		
	
	ini();
	dp();
	ans = c[n][m];
	for(int i = 1; i <= m / 2; i++) 
		swap(b[i],b[m - i + 1]);
	
	ini();
	dp();
	ans = min(ans,c[n][m]);
	printf("%d",ans);
	return ;

int main() 
	init();
	solve();
	return 0;